(a) What is the momentum of a garbage truck that is 1.20×104 kg and is moving at 30.0 m/s? (b) At what speed would an 8.00-kg trash can have the same momentum as the truck?
Solution:
Linear momentum (momentum for brevity) is defined as the product of a system’s mass multiplied by its velocity. In symbols, linear momentum \textbf{p} is defined as
\textbf{p}=m \textbf{v},where m is the mass of the system and \textbf{v} is its velocity.
Part A. The momentum of the garbage truck
The garbage truck has the following quantities given:
\begin{align*}
m_{\text{truck}} & = 1.20 \times 10^{4}\ \text{kg} \ \\
\textbf{v}_{\text{truck}} & = 30.0\ \text{m}/\text{s}
\end{align*}Substitute these values in the formula of momentum to solve for the momentum of the truck.
\begin{align*}
\textbf{p} & = m \textbf{v} \\
\textbf{p} & = \left( 1.20 \times 10^{4}\ \text{kg} \right) \left( 30.0\ \text{m}/\text{s} \right) \\
\textbf{p} & = 360000\ \text{kg} \cdot \text{m}/\text{s} \\
\textbf{p} & = 3.60 \times 10^{5}\ \text{kg} \cdot \text{m}/\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}Part B. The Speed of a Trash to have the same Momentum as the Truck
The trash has the following properties:
\begin{align*}
m & = 8.00\ \text{kg} \\
\textbf{p} & = 3.60 \times 10^{5}\ \text{kg} \cdot \text{m}/\text{s}
\end{align*}We now use the formula for momentum to solve for the unknown velocity.
\begin{align*}
\textbf{p} & = m \textbf{v} \\
\textbf{v} & = \frac{\textbf{p}}{m} \\
\textbf{v} & = \frac{3.60 \times 10^{5}\ \text{kg} \cdot \text{m}/\text{s}}{8.00\ \text{kg}} \\
\textbf{v} & = 45000\ \text{m}/\text{s} \\
\textbf{v} & = 4.5 \times 10^{4}\ \text{m}/\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}